Recent content by iamlorde

So, I would think, (b) P(X > 10) = [!not](everything up to and including 10) = 1-0.9994 = 0.0006 (d) P(6 ≤ X ≤ 11) : X is between 6 and 11, AND it includes them both, so I choose one above 11, so 12 which is: 1.0000. Then I choose one below 6, so 5, so that I can include 6. Hence: 0.8042...

Question is as follows: (a) = 0.4114 is the answer. Yet all I see from this answer is that X is simple equal to "0.4114". If it is "X ≤ 3" shouldn't "0.2061", "0.0692", and "0.0115" contribute to the answer somehow because they are "<" smaller than 3? I feel like I may be missing a...

I'm sorry but I still can't get the answer. My logic is as follow: P[(passwords with exactly 1 integer) OR (passwords with exactly 2 integer)] Therefore: [A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[A-Za-z]*[0-9] or [52]*[52]*[52]*[52]*[52]*[52]*[52]*[10] so: 52^7 * 10 over 62^8...

The following is my visualization of the problem (c). At this point I'm acknowledging that set (Password contains exactly 1 or 2 integers) is exclusive from set A & B. Yet I do not know how to quantitatively approach this problem. Thank you.

Jameson, thank you for your reply. I hadn't given the rest because I had answered them already. Here is the entirety. (a) $$ P(A \cup B) = \frac{{52}^{8}}{{62}^{8}}\ + \frac{{10}^{8}}{{62}^{8}}\ = 0.2448 $$ (b) $$ P(A' \cup B) = \frac{{62}^{8} - {52}^{8}}{{62}^{8}}\ = 0.7552 $$

Greetings! Can someone offer help? I proceeded like so: $$ \frac{{52}^{7}\cdot10}{{62}^{8}}\ + \frac{{52}^{6}\cdot{10}^{2}}{{62}^{8}}\ = 0.0561 $$ The answer is supposed to be: 0.630, but I have no idea how to reach it.